Kindly refer to the figure, AB is the tangent to the circle with cente

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GMATBusters’ Quant Quiz Question -7
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Kindly refer to the figure, AB is the tangent to the circle with center O, and angle AOB is 90 degrees. If OA =4 cm, and OB = 3cm. What is the radius of the circle?
A. 1.6
B. 2.1
C. 2.4
D. 3.2
E. 3.5

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Official Answer

C

gurmukh · Answer

1/2×3×4=1/2×5×r
r =2.4

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ashwini2k6jha · Answer

answer C
Let D be point radius from O touches tangent
triangle OAD and triangle OAB are simmilar
hence corresponding ratio of sides equal
radius/4=3/5
radius=2.4

uc26 · Answer

If we join the center of the circle with the point of tangency, we get a right angle. Let the point of tangency be T and the line be OT. OT is the radius of the circle.
We can see the AB=5 (3-4-5 pythagorean triplet)
Area of triangle AOB= 1/2 * AO* AB= 1/2* 3*4= 6
Also, Area= 1/2 * OT* AB
so 6= 1/2 * OT* 5
OT= 12/5= 2.4.
Answer: C

Archit3110 · Answer

given OB is 3 cm so radius of circle has to be < 3cm D &E are out 
now for 
∆ AOB ; side AB = 5 AO = 4 and BO = 3
∆ AOD OD = r and AO = 4
OD/OA= OB/AB 
=> r/4 = 3/5
=> r= 12/5 ; 2.4
OPTION C

Kindly refer to the figure, AB is the tangent to the circle with center O, and angle AOB is 90 degrees. If OA =4 cm, and OB = 3cm. What is the radius of the circle?
A. 1.6
B. 2.1
C. 2.4
D. 3.2
E. 3.5

Kinshook · Answer

Given: Kindly refer to the figure, AB is the tangent to the circle with center O, and angle AOB is 90 degrees.
Asked: If OA =4 cm, and OB = 3cm. What is the radius of the circle?

Attachment:

Screenshot 2020-05-24 at 8.52.15 AM.png [ 27.06 KiB | Viewed 3968 times ]

Since \(\triangle\) AOB & \(\triangle\) AOD are similar; Angle A & \(90^0\) are common.
\(\frac{OD}{AO} = \frac{r}{4} = \frac{OB}{AB} = \frac{3}{5}\)
Radius of circle = \(r = \frac{12}{5} = 2.4\)

IMO C

MayankSingh · Answer

IMO C

AB is the tangent to the circle with center O, 
OA =4 cm, and OB = 3cm.

Lets Draw radius OD from O at point of contact of tangent to circle.(Shown in figure)
Tangent is perpendicular to radius at point of contact.
So ∠ ADO= 90

In triangle AOB & ADO
∠AOB=∠ADO=90
∠OAB=∠DAO (Common)
Both triangles are similar
OB/AB= OD/AO
OD= OB x AO/AB = 3 X 4 /5 = 2.4

C. 2.4

tyson1990 · Answer

In triangle AOB, <AOB = 90 deg
OA = 4, OB = 3
using rule of AB^2 = OA^2 + OB^2
AB = 5

Now radius of circle is perpendicular to tangent
Hence considering rule of triangle area

1/2 X OA X OB = 1/2 X R X AB
R = (OA X OB) / AB = 3X4/5 = 2.4

shashwat91 · Answer

ab^2 = 4^2+3^2
ab=5
area ABO(considering base BO) = 1/2*3*4=6
area ABO(considering base AB) = 1/2*5*h=6 (h=point where circle and ab intersect/meet)
therefore h=12/5
and h=r
so r= 2.4(ans)

karan12345 · Answer

Area of triangle AOB =  \frac{1}{2} * base * height   
 = \frac{1}{2}  * 3 * 4 = 6 sq. cm

We know , radius of circle which joins centre to tangent makes right angle with circumference.

AB =  \sqrt{AO^2 + OB^2}  =  \sqrt{25}  = 5 cm (by pythagoras theorem)
 
Also, Area of triangle AOB =  \frac{1}{2}  * AB * radius of circle
=> 6 = 2.5 * radius of circle
=> Radius of circle =  \frac{6}{2.5}  = 2.4 cm

Hence, OA is C.

Codebug4it · Answer

Answer: OPTION C - 2.4

As AB is a tangent and radius of a circle is always perpendicular on a tangent  , hence we can draw a perpendicular from O to AB, as OD.

Using Pythagoras theorem,

AB ^ 2 = (OA) ^ 2 + (OB) ^ 2  = (4 ^ 2) + (3 ^ 2)
=>  AB = 5

Area of traingle = 1/2 * (base * height)

As triangle OAB is a right triangle and has two perpendiculars so,

=>  1/2 * (OA * OB) = 1/2 * (AB * OD) 
 =>  1/2 * 4 * 3 = 1/2 * 5 * OD 
 =>  OD = 12 / 5  =  2.4

Crank513 · Answer

Let's say that the points where the tangent is touching the circle be x. 
Angle OXB = 90, and angle XBO is 45. Sin 45 = perpendicular/hypotenuse. 
Hypotenuse is 3, so 1/√2 = ox/3, OX = 3/√2 
Which is 2.08 so 2.1

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urshi · Answer

Answer : Option C ( 2.4cm )

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Hulikunte · Answer

C .

Let point of intersection of circle with Tangent AB be ' D'. 
Also, AB = 5 for Right Angle Triangle AOB with legs OA = 4 & OB = 3

Area of Triangle AOB = 1/2 x OB x OA = 1/2 x AB x OD ( For Base AB , OD is the altitude )
Radius = OD = 12 /5 = 2.4 cm

adityasuresh · Answer

Equate the areas.
1/2*4*3 =6
Consider 5 as the base and draw a radius perpendicular to the base 5
Equate the areas.

Kindly refer to the figure, AB is the tangent to the circle with cente

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